$\dfrac{d}{dx}\left(\dfrac{x^2-x+5}{4x-1}\right)=$
Answer: $\dfrac{x^2-x+5}{4x-1}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( x 2 − x + 5 4 x − 1 ) = ( 4 x − 1 ) d d x ( x 2 − x + 5 ) − ( x 2 − x + 5 ) d d x ( 4 x − 1 ) ( 4 x − 1 ) 2 The quotient rule = ( 4 x − 1 ) ( 2 x − 1 ) − ( x 2 − x + 5 ) ( 4 ) ( 4 x − 1 ) 2 Differentiate ( x 2 − x + 5 ) & ( 4 x − 1 ) = 8 x 2 − 4 x − 2 x + 1 − 4 x 2 + 4 x − 20 ( 4 x − 1 ) 2 Expand = 4 x 2 − 2 x − 19 ( 4 x − 1 ) 2 \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{x^2-x+5}{4x-1}\right) \\\\ &=\dfrac{(4x-1)\dfrac{d}{dx}(x^2-x+5)-(x^2-x+5)\dfrac{d}{dx}(4x-1)}{(4x-1)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(4x-1)(2x-1)-(x^2-x+5)(4)}{(4x-1)^2} \gray{\text{Differentiate }(x^2-x+5)\text{ & }(4x-1)} \\\\ &=\dfrac{8x^2-4x-2x+1-4x^2+4x-20}{(4x-1)^2} \gray{\text{Expand}} \\\\ &=\dfrac{4x^2-2x-19}{(4x-1)^2} \end{aligned} In conclusion, $\dfrac{d}{dx}\left(\dfrac{x^2-x+5}{4x-1}\right)=\dfrac{4x^2-2x-19}{(4x-1)^2}$, or any other equivalent form.